What is the area of the region between the graphs of $f(x)=x^2+12x$ and $g(x)=3x^2+10$ from $x=1$ to $x=4$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $45$ (Choice B) B $77$ (Choice C) C $\dfrac{64}{3}$ (Choice D) D $18$
Solution: Visualizing the area We sketch the graphs of $f$ and $g$ first. ${1}$ ${2}$ ${3}$ ${4}$ ${20}$ ${40}$ ${60}$ $f$ $g$ $y$ $x$ From the graph, it appears that $f(x)\ge g(x)$ between $x=1$ and $x=4$. From this we are looking to evaluate: $ \int_{1}^{4}\left( f(x)-g(x) \right)\,dx$ Evaluating the definite integral $\begin{aligned} &\phantom{=} \int_{1}^{4} \left( x^2+12x- (3x^2+10) \right) \,dx \\\\ &= \int_{1}^{4} \left( -2x^2+12x-10 \right) \,dx \\\\ &= -\dfrac{2x^3}{3}+6x^2-10x~\Bigg|_{1}^{4} \\\\ &= \left( -\dfrac{128}{3}+96-40 \right) -\left( -\dfrac{2}{3}+6-10 \right)\\\\ &= 18 \end{aligned}$ Answer The area is $18$ square units.